Ffwythiant Dwysedd Tebygolrwydd f(x)
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Transcript Ffwythiant Dwysedd Tebygolrwydd f(x)
Ffwythiant Dwysedd Tebygolrwydd f(x)
Probability Density Function f(x)
Os oes gan hapnewidyn di-dor werth posib rhwng a a b,
gallwn ddarlunio sut y rhennir un uned o debygolrwydd
rhwng y gwerthoedd yma mewn graff o f(x).
If a continuous random variable has a value between a and b, we can
show how one unit of probability is distributed in a graph of f(x).
f(x)
=1
a
b
x
Gan fod rhaid i gyfanswm yr arwynebedd o dan y gromlin
fod yn hafal i 1 cyfan (cyfanswm tebygolrwydd), mae
The area under the curve (the total probability) must be equal to 1,
therefore
b
f ( x)dx 1
a
I ddarganfod P(c ≤ x ≤ d)
To calculate P(c ≤ x ≤ d)
f(x)
d
a
c d
b
x
f ( x ) dx
c
Pan fo X yn ddi-dor, gellir newid y symbol ≤ a rhoi < yn ei le
fel bod
P(c ≤ X ≤ d) = P(c < X < d) = P(c ≤ X < d) = P(c < X ≤ d)
When X is continuous, the ≤ symbol can be replaced with < so that
P(c ≤ X ≤ d) = P(c < X < d) = P(c ≤ X < d) = P(c < X ≤ d)
Enghraifft - Example
Dosrennir yr hapnewidyn di-dor X gyda ffwythiant dwysedd
tebygolrwydd f a roddir gan
X is a continuous random variable with a probability density function
f(x) = kx(4-x)
Darganfyddwch werth
Find the value of
a) k
b) P(X ≤ 3)
c) P(0 < X < 1 | X ≤ 3)
ar gyfer/for 0 ≤ x ≤ 4
4
a)
kx(4 x)dx 1
0
4
k ( 4 x x 2 ) dx 1
0
4
4x
x
k
1
3 0
2
2
3
3
3
4
0
2
2
k 2(4) 2(0) 1
3
3
64
k 32 1
3
32 k
1
3
3
k
32
3
b) P(X ≤ 3) =
3
0 32 x(4 x)dx
3
3
2
(
4
x
x
)dx
32 0
3
3 2 x
2 x
32
3 0
3
3
3
3
3
0
2
2
2(3) 2(0)
32
3
3
3
3 9
18 9 0
32
32
27
32
c) P(0 < X < 1 | X ≤ 3) = P((0 < X < 1) (X ≤ 3))
P(X ≤ 3)
P(A|B) = P(A B)
P(B)
P((0 < X < 1) (X ≤ 3)) = P(0 < X < 1)
1
P(0 < X < 1) =
3
2
(
4
x
x
)dx
32 0
1
3
2 1 3
2
x
x
32
3
0
3
1 3
1 3
2
2
2(1) (1) 2(0) (0)
32
3
3
3 5 5
32 3 32
5 27
P(0 < X < 1 | X ≤ 3) =
32 32
5
27
Ymarfer/Exercise 1.1
Mathemateg - Ystadegaeth Uned S2 – CBAC
Mathematics Statistics Unit S2 - WJEC
Gwaith Cartref/Homework 10